Calculate the leakage inductance for the windings in the pot core shown in Figure 6.19 with the dimensions given in Table 6.2.

Question

Accepted Answer

The MLT is based on the average radius of the coils:

MLT = 2 × π × (9.86 + 13.04)/2 = 71.94 mm

The breadth of the winding is b = 13.04 - 9.86 = 3.18 mm

The leakage inductance is:

[latex]L_{l 1}+L_{l 2}=\frac{\mu_0 N_p^2 MLT b}{3 w}=\frac{\left(4 imes 10^{-7} ight)(20)^2\left(71.94 imes 10^{-3} ight)\left(3.18 imes 10^{-3} ight)}{(3)\left(17.6 imes 10^{-3} ight)} imes 10^6=2.2 \mu H[/latex]

The result in Equation 6.81

[latex]L_{l 1}+L_{l 2}=\frac{\mu_0 N_p^2 MLT b}{3 w}[/latex] (6.81)

was based on the low frequency distribution of the H field in Figure 6.4 and that, in turn, was based on the low frequency current distribution, i.e. constant current density across each layer. However, at high frequency, the proximity effect redistributes the current, as illustrated in Figure 6.15.

Performing the integration in Equation 6.79

[latex]W_m=\frac{1}{2} \mu_0 \int_{ ext {volume }} H^2 d V[/latex] (6.79)

on the high-frequency distribution on Figure 6.15 results in lower energy and, consequently, lower leakage inductance. Remember, the real part of the expression in Equation 6.47

[latex]\frac{R_{ ac }}{R_{ dc }}=\Re\left(\Delta(1+j)\left[\operatorname{coth}(\Delta(1+j))+\frac{2\left(p^2-1 ight)}{3} anh \left(\frac{\Delta}{2}(1+j) ight) ight] ight)=\Delta\left[\frac{\sinh 2 \Delta+\sin 2 \Delta}{\cosh 2 \Delta-\cos 2 \Delta}+\frac{2\left(p^2-1 ight)}{3} \frac{\sinh \Delta-\sin \Delta}{\cosh \Delta+\cos \Delta} ight][/latex] (6.47)

represents the loss as predicted by Dowell’s formula. The imaginary part of the expression represents the leakage reactance, and this can be easily deduced:

[latex]\frac{L_l ac }{L_l}=\frac{3}{2 p^2 \Delta^2} \operatorname{Im}\left(\Delta(1+j)\left[\operatorname{coth}\left(\Delta(1+j) +\frac{2\left(p^2-1 ight)}{3} anh \left(\frac{\Delta}{2}(1+j) ight) ight] ight) ight. =\frac{3}{2 p^2 \Delta}\left[\frac{\sinh 2 \Delta-\sin 2 \Delta}{\cosh 2 \Delta-\cos 2 \Delta}+\frac{2\left(p^2-1 ight)}{3} \frac{\sinh \Delta+\sin \Delta}{\cosh \Delta+\cos \Delta} ight][/latex] (6.82)

This is plotted in Figure 6.20. There is a fall off in leakage inductance for Δ > 1.

It is common practice to depend on leakage flux to provide the resonant inductor in resonant power supplies. The dramatic fall-off when the skin depth is comparable to the layer thickness must be considered in the design.

Table 6.2 Coil dimensions
Coil 1:	Number of turns	20
	Inside radius	9.86 mm
	Outside radius	11.51 mm
	Width	17.6 mm
Coil 2:	Number of turns	20
	Inside radius	11.39 mm
	Outside radius	13.04 mm
	Width	17.6 mm

Question 6.5: Calculate the leakage inductance for the windings in the pot...

Related Answered Questions

Push-pull Converter Calculate the optimum thickness for the winding in the push-pull converter in Example 5.3. ...

Verified Answer:

Repeat Example 6.2 using the approximation in Equation 6.75 to find the optimum value of Δ. ...

Verified Answer:

Calculate the ratio of effective resistance to DC resistance Reff/Rdc in a copper foil of thickness d carrying the pulsed current waveform shown in Figure 6.9. This waveform is often encountered in switch mode power supplies. ...

Verified Answer:

Calculate the ratio of effective resistance to DC resistance Reff/Rdc in a copper foil of thickness d carrying the pulsed current waveform shown in Figure 6.12. ...

Verified Answer: